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OPOSICIÓN MAGISTERIO EDUCACIÓN INFANTIL GALICIA
Vectors and Forces Your average speed for the walk is: total distance÷total time = 8÷4 = 2ms-1 (since it takes 5÷2 = 2.5s walking North and 3 ÷2 = 1.5s walking South).
Vector quantities are commonly encountered in physics. This Factsheet will explain: ♦ the difference between a vector and a scalar, giving common examples of each ♦ how to add and subtract vectors ♦ how to resolve vectors ♦ how to apply this to forces and particles in equilibrium
However, your average velocity is: total displacement ÷total time = 2m North ÷4 = 0.5ms-1 North Tip: Always be very careful not to use “speed” when you mean “velocity” and vice versa – ask yourself first whether it has a direction.
1. What is a vector?
A vector is a quantity that has both magnitude and direction.
Typical Exam Question a) State the difference between scalar and vector quantities b) Give two examples of a vector quantity.
A scalar just has a magnitude – it is just a (positive or negative) number.
a) a vector has direction ü a scalar does not b) any two examples from vector column in table 1 üü
A vector can be represented by an arrow – the length of the arrow represents the magnitude of the vector, and the direction in which the arrow is pointing shows you the vector’s direction.
2. Adding Vectors Anything that has a direction as well as a size will be a vector – for example, if you tell someone that your house is 200m East of the chip shop, you are representing the position of your house by a vector – its magnitude is 200m and its direction is East. If, instead, you just said that your house was 200m away from the chip shop, you are using a scalar quantity (distance). You will notice that the scalar is not so useful as the vector in telling someone where your house is!
Imagine driving 40 miles East then 30 miles North. Your path would look like: finish 30 miles
Table 1 shows some common examples of scalar and vector quantities.
Table 1. Scalars and Vectors Scalars Distance Speed Temperature Energy Power Pressure Mass
40 miles How far away from your starting point have you ended up? It certainly isn’t 70 miles. If you draw the other side of the right-angled triangle, and use Pythagoras’ Theorem, you will find it is 50 miles.
Vectors Displacement Velocity Acceleration Force Momentum Torque/Moment Impulse
To describe the position at which you ended up, you’d say it was 50 miles from your starting point, but you’d also need to say in what direction. It won’t be a direction like Northeast (since that would mean going the same distance North and East), so we’d have to give the direction in terms of an angle.
Exam Hint: Exam questions often require you to explain the difference between a scalar and a vector, and to indicate whether a particular quantity is a vector or scalar. If you think you might find it hard to work this out in an exam, make sure you learn the common examples.
You might choose to find the angle marked α in the diagram below. finish
It is particularly important to understand the distance between distance and displacement, and between speed and velocity (and hence average speed and average velocity):
Suppose you walk 5m North, then 3m South. The total distance you have travelled is obviously 8m. However, your final displacement – which just means where you end up relative to where you started – is 2m North.
αα 40 miles
Suppose you were walking at a steady 2 ms-1. Then your speed was constant throughout – it was 2 ms-1. However, your velocity was not constant, since to begin with you were travelling at 2ms-1 North, then you changed to 2ms-1 South.
Using trigonometry (see the Factsheet Maths for Physics: Trigonometry if you need help on this), we find α = tan-1(0.75) = 36.9o.
Vectors and Forces But it is more conventional to give directions as a bearing – measured clockwise from North. So we’d need the angle marked β in the diagram below: finish
Tip: When you are using the cosine rule, take care with “BIDMAS” – you must work out 2bc cosA, and subtract the answer from b2 + c2. Also, always check that your answer sounds sensible – if it does not, you may have forgotten to square root.
αα 40 miles
We can see that α + β = 90o, so β = 53.1o. Now we need to find the direction. Since in the question angles are given to the horizontal, we should give the answer in that way. So we want angle α shown below:
We can now describe the final displacement as: 50 miles at a bearing of 53.1o
We use this instead of the We use this instead of the rounded value to avoid rounded value to avoid roundingerrors. errors. rounding
The example above was an example of vector addition. We added two displacements (40 miles East and 30 miles North) to find a resultant diplacement (50 miles at a bearing of 53.1o). In this case, we found the resultant by calculation (using Pythagoras’ Theorem and trigonometry). It could also have been found by scale drawing.
√9100 50 α
Tip: In any problem of this type, you MUST draw a diagram – even if it is only a sketch.
sinB sinC = ; = a b c A=120o, a = √9100, b = 60, c = 50, C = α Use the sine rule:
Finding the resultant of two vectors To find the resultant of two vectors: ♦ Draw one of the vectors ♦ Draw the other vector starting at the end of the first one. ♦ Draw an arrow from the start of the first vector to the end of the second one – this represents the resultant ♦ Use calculation or accurate measurement (if you are told to use scale drawings) to find the length (magnitude) of the resultant. ♦ Find the direction of the resultant
We leave out the part involving sinB, since we are not interested in it, and it is not useful. So we have: sin120 sinα = 50 9100 sin120 × 50 = sinα 9100 0.45392... = sinα 27o = α (nearest degree)
Exam Hint: If you are asked to find the resultant, you must give its direction as well as its length if you want full marks
So the resultant has magnitude 95cm and is at an angle of 27o above the horizontal.
Example 1. Vector a is of magnitude 60cm and acts horizontally. Vector b is of magnitude 50cm and acts at 60o above the horizontal. Find the resultant of vectors a and b.
Parallelogram rule of vector addition Let us consider adding two vectors, u and v, to give their resultant, R. R is the diagonal of the parallelogram whose sides are u and v. This gives another visual way to think about vector addition.
Tip: You can write the sine rule with all the sines on the top, or with all the sines on the bottom. If you are trying to find an angle, have them on the top, and if you are trying to find a side, have them on the bottom.
If you are using calculation, you may need to use the sine and cosine rules as well as trigonometry in normal right-angled triangles. If you do not like this sort of trigonometry, you may prefer to use the alternative method given at the end of section 3 in this Factsheet.
First find the magnitude of the resultant; to do this we use this triangle:
The parallelogram can also be used to subtract vectors - to find u - v, you'd draw the parallelogram using vectors u and -v:
o 0 120
u R -v
u Note that for calculation purposes, the parallelogram works exactly the same way as the vector triangle.
Comes from 180o – 60o
Vectors and Forces
Example 2. A vector of magnitude 10 is inclined at 30o above the horizontal. Find the horizontal and vertical components of the vector.
Resultant of more than 2 vectors To find the resultant of more than 2 vectors, we draw a vector polygon. This is very similar to finding the resultant of 2 vectors, except that you draw the third vector after the second vector, and so on. The diagram shows the vector addition of vectors a, b, c and d, and their resultant, R.
h , so h = 10 × cos30o = 8.66 10 v sin30o = , so v = 10 × sin30o = 5 10 cos30o =
In fact, you can save time working out the trigonometry by remembering the following diagram showing the vector a and its components:
If you need to find the resultant of 3 or more vectors by calculation, it is best to use the method described at the end of section 3.
asinθ a sinθ
Special case: If the vector polygon closes (so the end of the last vector coincides with the start of the first one), then the resultant is zero.
The following example shows how to apply this. Example 3. The diagram below shows a vector V and the line AB.
Tip: It does not matter in which order you add vectors, so if it makes your diagram or calculation easier to put them in a particular order, go ahead! V B
3. Resolving Resolving a vector involves writing it as the sum of other vectors – it’s like resolving in reverse. For example, the vector a shown below can be written as the sum of a horizontal vector (h) and a vertical vector (v).
A The angle between V and AB is 20o. Find the components of V parallel and perpendicular to AB, given that the magnitude of V is 15. Our triangle is as shown:
15 v 20o
h The seperate vectors that the original is resolved into are called components – in the above example, h is the horizontal component of a and v is the vertical component of a.
By comparison with the diagram above, we get:
Note that there are many other ways we could resolve vector a – we choose the most convenient way in each situation (see section 4 for some examples of this). It is, however, always best to resolve a vector into two perpendicular components.
Exam Hint: In practical problems, the two directions are usually horizontally and vertically, or if an inclined plane is involved, along and perpendicular to the plane.
o o 15 cos20 15cos20
So the component parallel to AB is 15cos20o = 14 (2 SF) The component perpendicular to AB is 15sin20o = 5.1 (2SF) Typical Exam Question Bill travels 10km North-east and then 12km due East (a) Draw a vector diagram showing Bill’s route.  (b) Calculate, without the use of a scale diagram, Bill’s resultant displacement in components East and North.  12km ü (a)
Calculating components To find the components of a vector in a pair of perpendicular directions, we will be using a right-angled triangle. ♦ Take the vector as the hypotenuse of the triangle ♦ Take the directions you want to resolve in as the other two sides ♦ Use trigonometry to work out the size of the components.
(b) E: 10cos45 ü+ 12ü = 19km.
N: 10sin45 = 7.1kmü
Vectors and Forces
Using components to find the resultant For problems involving many vectors, the following provides a fail-safe Problems involving equilibrium often require you to use the fact that the body is in equilibrium to find unknown forces. The procedure here is: method to find the resultant force: ♦ First read the question carefully to check whether the particle is in equilibrium ♦ Choose two sensible perpendicular directions – horizontal and vertical ♦ Draw a diagram showing all the forces are often a good idea. ♦ Resolve all forces in two perpendicular directions ♦ Resolve every vector involved in these two directions (you could put them in a table to aid clarity and ensure you have not missed one out) ♦ Find the total of the components in each direction, and equate it to 0. ♦ Add up all the components in one direction (say horizontal), which ♦ Solve your equations to find any unknown forces. gives the resultant force in that direction. Example 5. A box of mass 10kg is being towed at constant velocity along Repeat for all the components in the other direction. o You are now left with two perpendicular vectors. Find the resultant of rough horizontal ground by a rope inclined at 30 to the horizontal. The tension in the rope is 40N. Find: these two vectors using Pythagoras and basic trigonometry. a) the frictional force exerted by the ground on the box b) the normal reaction force exerted by the ground on the box. Exam Hint: A common mistake is to resolve correctly, but ignore the Take g = 9.8ms-2 direction of the component – for example, 6 units upwards is not the same as 6 units downwards! To avoid confusion, decide at the beginning which We know the box is in equilibrium because it is moving with constant velocity direction to take as positive.
Normal reaction N Tension = 40N
Example 4. Find the resultant of the vectors shown below. 30 6
8 weight = 10g We will take to the left, and upwards as positive Vector 9
Resolve horizontally and vertically (taking upwards and left as positive), and equate to 0:
Horizontal compt 9
Vertical compt 0
6 cos30o -4cos50o
40cos30o – F = 0 40sin30o + N – 10g = 0
From the first equation, we get F = 40cos30o = 35N (2 SF) From the second equation, we get N = 10g – 40sin30o = 78N
The angles 50o and 60o come from using angles on a straight line = 180o
Example 6. A box of mass 10kg is at rest on a rough plane inclined at 5o to the horizontal. Find the normal reaction and the frictional force exerted by the plane on the box. Take g = 9.8 ms-2
Tip: Do not actually work out the sines and cosines yet, to avoid rounding errors or copying errors.
N So total of horizontal components = 9 + 6cos30o – 4cos50o – 8cos60o = 7.625... total of vertical components = 0 + 6sin30o + 4sin50o – 8sin60o = -1.864...
So to find overall resultant: By Pythagoras, magnitude of R = √(7.6252 + 1.8642) =7.8 (2 SF) α = tan-1(1.864 ÷ 7.625) = 14o below the horizontal (2 SF)
The box is in equilibrium as it is at rest. Note that friction must act up the plane, since the box will be “trying” to slide down.
4. Application to forces and equilibrium
To find the resultant of a number of forces, use the methods described When there is an inclined plane involved, it is best to resolve parallel and above. perpendicular to the plane. A body is in equilibrium if there is no resultant force (and no resultant o torque – see Factsheet 4 Moments and Equilibrium) on it. In the examples The weight of the box acts at an angle of 85 to the plane (from angles in a considered in this Factsheet, there will never be a resultant torque, so we triangle). So we have: will only have to use that the resultant force is zero. Along the plane: F – 10gcos85o = 0 ⇒ F = 8.5N (2SF) N – 10gsin85o = 0 ⇒ N = 98N (2SF) Any body is in equilibrium if it is at rest, or moving with constant velocity Perpendicular to the plane: (not just a constant speed) – see Factsheet 12 Applying Newton’s Laws for more details on this.
Vectors and Forces Typical Exam Question A body of mass 5.0kg is pulled up a rough plane, which is inclined at 30o to the horizontal, by the application of a constant force of 50N, which acts parallel to the plane. Take g = 9.8 ms-2 (a) Draw a free body diagram for the body showing the normal reaction R, frictional force F, weight W and the applied force A. (b) When the arrangement is in equilibrium, what are the values of the normal reaction force R and the frictional force F? 
Exam Workshop This is a typical poor student’s answer to an exam question. The comments explain what is wrong with the answers and how they can be improved. The examiner’s answer is given below. A boat travels at 6.8ms-1 South-easterly towards a harbour, which is 10km away. Once the boat reaches harbour, the passengers get in a car and drive due North at 80kmhr-1 for 15 minutes. Calculate the: (a) total distance travelled.  d = s × t = 80 × ¼ = 20 + 10 = 30 ü
1 mark deducted for omission of units. Although it would not lose marks, the candidate should avoid writing 80 × ¼ =...= 30, since it is not mathematically correct and could lead to confusion
W ü (b) Resolve forces perpendicular to the plane: R = Wcos30o ü so R = 42.4 ≈42Nü Resolve forces parallel to the plane: F + Wsin30o = 50 ü so F= 21.7 ≈ 22N ü
(b) total displacement. 202 – 102 = 300. √300 = 17.3 km
Questions 1. a) Explain the difference between a vector and a scalar b) Indicate whether each of the following is a scalar or a vector: Density Momentum Electrical resistance Distance Acceleration
0/5 Candidate has attempted to treat this as a right-angled triangle, when it is not – either the cosine rule or resolving into components should have been used. Candidate has also not given the direction
2. Find the resultant of each of the following: a) A force of 5N acting horizontally to the left and a force of 8N acting vertically upwards b) A force of 10N acting vertically upwards and a force of 3N acting at 20o to the upward vertical. c) A force of 5N acting at 10o below the horizontal and a force of 5N acting at 10o above the horizontal d) A force of 6N acting horizontally to the left, a force of 8N acting vertically upwards and a force of 2N acting horizontally to the right.
(c) average speed. 80kmh-1 = 80 × 1000/3600 = 22.2ms-1 So average speed = (22.2 + 6.8) ÷ 2 = 14.5ms-1
3. A body is in equilibrium. Which of the following must be true? There may be more than one correct answer. a) the body is stationary b) the polygon of forces for the body is closed c) there are no forces acting on the body.
Candidate has not used the correct definition of average speed – it is total distance/ total time, not the average of the individual speeds. (d) magnitude of the average velocity. 14.5ms-1
4. Find, by calculation, the resultant of the forces shown below : 20N
o o 1 0 0100 o 120
Poor exam technique – the candidate should appreciate that 2 marks will not be awarded for simply writing down the same answer again. Although the candidate’s answer for total displacement was incorrect, credit would have been awarded if it had been used correctly to find average velocity.
5. A smooth bead of mass 10g is threaded onto a thin piece of string of length 2m. The ends of the string are fastened to the ceiling so that they are at the same horizontal level as each other and are 1.6m apart. a) Draw a diagram to show the forces acting on the bead. b) Explain why the bead only rests in equilibrium at the midpoint of the string. c) Find the tension in the string, taking g = 10ms-2
Examiner's Answers (a) Total distance = 10 + (80 × ¼) ü = 30km ü (b) Total displacement = 10km SE + 20km N East = 10 cos45 North = -10 sin45 + 20. = (7.1, 12.9) km ü Magnitude is |R|=√(7.12+12.92) ü=14.7 km Direction is given by θ = tan-1(12.9 / 7.1) ü= 61.2o Displacement vector = 15km, ü 61° N of E (or bearing 029) ü (c) Average speed = total distance / total time = 30 km/ (25 + 15) minutes ü = 0.75 km / minute = 45 km / hr ü (d) Magnitude of average velocity = (magnitude of total displacement) / total time = (14.7 km)/ ( 2/3 hr) ü = 22 km / hr ü
Answers 1. a) A vector has magnitude and direction; a scalar has magnitude only b) scalar, vector, scalar, scalar, vector 2. a) magnitude 9.4N, at 58o above horizontal b) magnitude 13N, at 4.6o to upward vertical c) 9.8N acting horizontally d) 8.9N acting at 63o above (leftward) horizontal 3. b) only 4. 13N at 11o above rightward horizontal 5. a) T T
b) If the bead were not at the midpoint, the angles made by the two pieces of string would not be equal. If the angles were not equal, there would be a net horizontal force on the bead,since the horizontal components of the tensions would not balance, so the bead would not be in equilibrium. c) 0.083N (2SF)
Acknowledgements: This Physics Factsheet was researched and written by Cath Brown. The Curriculum Press,Unit 305B, The Big Peg,120 Vyse Street, Birmingham, B18 6NF. Physics Factsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publisher. ISSN 1351-5136